A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity u = Kt where K = 4.4 ms^-2 , the new time period will be (Take g = 10 ms^-2) 

Original time period is

$T_1=2\pi \sqrt{\frac{l}{g}}$                            (1)

When the pendulum is moving upwards, the effective value of g is

$g_{eff} = g + a$

where a is the acceleration of the pendulum which is given by

$\begin{array}{r}a=\frac{dv}{dt}=\frac{d}{dt}\left(Kt\right)=K=4.4{\mathrm{ms}}^{-2}\\ \therefore g_{\text{eff }}=g+a=10+4.4=14.4{\mathrm{ms}}^{-2}\end{array}$

Therefore, the new time period is

$T_2=2\pi \sqrt{\frac{l}{g_{\text{eff }}}}$                        (2)

From (1) and (2), we get

$$\begin{gathered} \frac{T_2}{T_1}=\sqrt{\frac{g}{g_{\mathrm{eff}}}}=\sqrt{\frac{10}{14.4}}=\frac{1}{1.2} \\ or T_2=\frac{T_1}{1.2}=\frac{3}{1.2}=2.5\mathrm{s} \end{gathered}$$